Author Topic: Capacitive coupling  (Read 4422 times)

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Offline socratidion

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Capacitive coupling
« on: July 14, 2015, 09:28:56 am »
Very basic question. I understand that capacitors block DC; I understand that they allow through AC. I observe it on my oscilloscope. But I don't understand why, in a circuit powered by a 9V battery, a signal wave (usually I'm dealing with pulse waves) with a max value of 9V (almost) and a min value of 0V (or just over) is change by the coupling capacitor into a wave with a max value of 4.5V and a min value of –4.5 V. I just don't see where the DC is that's being taken away. To my mind, a voltage range of 9 to 0 is the same thing as a voltage range of +4.5 to –4.5, just a matter of redefining the ground you stand on.

Is there a mathematical explanation that might put me at ease – show the logic of the cap voltage on the negative side going from 1/2 Vin to – 1/2 Vin? Or is there some other horrific misunderstanding I'm in the grips of?
 

Offline Ian.M

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Re: Capacitive coupling
« Reply #1 on: July 14, 2015, 09:42:50 am »
The scope input divider chain that sets the input voltage sensitivity (Volts/div) is equivalent to a 1M resistor to ground.  This forces the average (mean) value of the voltage after the capacitor to be 0V (assuming no other connections apart from the scope probe), and any symmetrical waveform will therefore have equal positive and negative peaks after the capacitor.   If you apply an asymmetric waveform the output will no longer be centred on 0V but will have equal areas above and below, between the trace and the 0V level.
 

Offline tautech

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Re: Capacitive coupling
« Reply #2 on: July 14, 2015, 10:03:32 am »
Where it shines is when you have a small AC component on a much larger DC signal.
AC coupling removes/blocks the DC and allows you to examine the AC only at the attenuation setting you desire.
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Offline Zero999

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Re: Capacitive coupling
« Reply #3 on: July 14, 2015, 10:24:50 am »
A symmetrical waveform oscillating between 9V and 0V has an average voltage of 4.5V so if you removed the AC component, you'd get 4.5VDC. On the other hand, if you remove the DC component, the waveform will swing between +4.5V and -4.5V which has an average voltage of 0V.
 

Offline socratidion

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Re: Capacitive coupling
« Reply #4 on: July 14, 2015, 11:29:42 am »
Thank you for your responses.

In a sense I feel you're only restating my puzzlement in different terms. I can see that, yes, if one removes the average voltage of the signal, then you would get the effect I am observing. But what is the mechanism by which the capacitor removes the average voltage? How does it know what the average is?
 

Offline McBryce

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Re: Capacitive coupling
« Reply #5 on: July 14, 2015, 11:42:27 am »
But I don't understand why, in a circuit powered by a 9V battery, a signal wave (usually I'm dealing with pulse waves) with a max value of 9V (almost) and a min value of 0V (or just over) is change by the coupling capacitor into a wave with a max value of 4.5V and a min value of –4.5 V. I just don't see where the DC is that's being taken away.

It doesn't and can't. Could you possibly upload a picture of the circuit and where you believe the signal has changed from being 0 - 9V into a -4.5 - 4.5V signal.
The only way it can "turn into" a -4.5V to +4.5V signal is if you are measuring the signal relative to the midpoint between the 0V and 9V battery potential.

McBryce.
 

Offline Ian.M

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Re: Capacitive coupling
« Reply #6 on: July 14, 2015, 11:58:11 am »
McBryce seems to be a little confused.  Take a simple 50% duty cycle square wave signal with high level of 9V and low level of 0V (e.g. from a CMOS 555 running from a 9V battery), and connect it to a scope via a 1uF capacitor (+ lead towards the signal source if its an electrolytic).  Select DC coupling for the scope input, and you will get a 9V peak to peak square wave, centred about 0V exactly as you have observed.   No midpoint tap on the battery to connect to scope probe ground is required or wanted.
« Last Edit: July 14, 2015, 12:44:54 pm by Ian.M »
 

Online MarkF

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Re: Capacitive coupling
« Reply #7 on: July 14, 2015, 12:36:13 pm »
Here is a simple explanation of how a capacitor works. 
 

Offline McBryce

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Re: Capacitive coupling
« Reply #8 on: July 14, 2015, 12:45:23 pm »
McBryce seems to be a little confused.  Take a simple 50% duty cycle square wave signal with high level of 9V ad low level of 0V (e.g. from a CMOS 555 running from a 9V battery), and connect it to a scope via a 1uF capacitor (+ lead towards the signal source if its an electrolytic).  Select DC coupling for the scope input, and you will get a 9V peak to peak square wave, centered about 0V exactly as you have observed.   No midpoint tap on the battery to connect to scope probe ground is required or wanted.

Ah ok. I completely mis-understood what he meant... Time for more coffee :)

McBryce.
 

Offline Ian.M

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Re: Capacitive coupling
« Reply #9 on: July 14, 2015, 12:48:12 pm »
I should clarify that the scope probe ground would be connected to the 0V (common) terminal of the signal source. (i.e. 9V battery -v terminal).

Posting before coffee or after whisky is frequently embarrassing . . . 8)
 

Offline McBryce

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Re: Capacitive coupling
« Reply #10 on: July 14, 2015, 12:54:05 pm »
I automatically assumed he meant the capacitor was across the signal (between square wave and battery minus). The spoil sports have banned Whisky for breakfast here! :D

McBryce.
 

Offline John Coloccia

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Re: Capacitive coupling
« Reply #11 on: July 14, 2015, 01:21:15 pm »
Thank you for your responses.

In a sense I feel you're only restating my puzzlement in different terms. I can see that, yes, if one removes the average voltage of the signal, then you would get the effect I am observing. But what is the mechanism by which the capacitor removes the average voltage? How does it know what the average is?

Don't think of it in terms of removing anything.  Think of the capacitor in terms of it's frequency dependent impedance.  At DC (0 Hz), the impedance of the capacitor is infinite...i.e, it looks like an open circuit.

So what happens if you have 5VDC on the left side, and 2VDC on the right side of the capacitor?  The same thing that happens if you have an open circuit.  When you measure the right side to ground, you'll see 2VDC, when you measure the left side to ground, you'll measure 5VDC, and when you measure across, you'll measure 3VDC....just like an open circuit.

As the frequency increases, the impedance of the capacitor goes down.  So put a 100Hz, 1V peak to peak signal in on the left side.  Toss that on the scope and you'll see a 1V peak to peak, 100Hz signal with a 5V DC offset.  What do you see on the right side?  Well, it depends on the value of the capacitor. Possibly, you'll measure a .5V peak to peak signal with a 2VDC offset.  At 100MHz, perhaps you'll measure a .999V peak to peak signal at a 2V DC offset (I didn't calculate anything...I'm just making up numbers for illustration).

Thinking about it in terms of magically blocking DC while letting through AC will get you into all sorts of trouble.  The right way to think about it is in terms of it's impedance.

So to answer your specific question, the DC isn't being taken away.  When you first turn on your system, the DC goes into charging up the capacitor.  If you have a scope probe on the input side when the system first turns on, you'll see the DC offset rise until you get to steady state.  That's the cap charging up.

Now you have some 9V peak to peak signal.  Let's say you want to amplify that with an opamp, but you don't want to use a dual supply.  You can't just AC couple it and feed it in because it will clip when it hits the negative 0V rail.  You need to bias the opamp so that it's operating at, say, 10V.  No problem.  Between the cap and the opamp input, simply put some large resistor (1M, 470K...something like that) to some 10V bias voltage.  When you turn the system on, after a little time everything will reach steady state, and you'll end up with your 9V peak to peak signal riding on top of a 10V DC offset.

Now let's use it to filter a power supply.  Put a cap from positive to ground.  When you turn the system on, you'll see the voltage rise on the + side until it hits steady state.  Remember, it's a DC voltage you want, but when you first turn it on, it looks like an AC signal...something approximating the first half of a square wave.  :)  But after a bit, the cap charges up and you reach stead state.  Then there is some noise on the line, some AC signal riding on top of that.  You select your cap to be large enough so that the impedance at whatever noise frequencies you're expecting is very low.  Ideally, you want DC to look like an open circuit to ground, and any noise to look like a dead short to ground.  In real life, neither of these is ever achievable, but you just have to get good enough that the noise is attenuated sufficiently.

« Last Edit: July 14, 2015, 01:44:10 pm by John Coloccia »
 

Offline cyr

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Re: Capacitive coupling
« Reply #12 on: July 14, 2015, 02:20:03 pm »
I just don't see where the DC is that's being taken away. To my mind, a voltage range of 9 to 0 is the same thing as a voltage range of +4.5 to –4.5, just a matter of redefining the ground you stand on.

One thing to remember is that your scope becomes part of your circuit when you connect is. It becomes mainly like a large value resistor to ground. That's what special about the ground point you have chosen.

Now, imagine you have an steady 4.5V on the input side of the capacitor instead of the 0-9V square wave. Current will flow through the capacitor and the scope (and back to ground) until the voltage on the "output" side of the capacitor is at ground level, the capacitor will then be charged up to (-) 4.5V.

From a DC point of view the same thing is happening in you circuit, the capacitor gets charged up to the average input voltage. To be precise it gets charged and discharged a little bit with every high and low period, but you hardly notice that if the capacitance and switching frequency is high enough.
 

Offline TimFox

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Re: Capacitive coupling
« Reply #13 on: July 14, 2015, 07:08:09 pm »
The easiest way to understand a series capacitor is that it cannot pass direct current from one end to the other.
Therefore, if the "output" end of the capacitor terminates in a resistance, there will be zero DC into the resistor and therefore zero DC level of the voltage across the resistance.
Things get more interesting with non-resistive loads, such as a diode, but the basic rule of zero direct current through the capacitor is always true, unless the capacitor has a "leakage" current or parallel resistance.  The series resistance of the capacitor does not change this rule.
 

Offline socratidion

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Re: Capacitive coupling
« Reply #14 on: July 15, 2015, 09:13:33 am »
Once again thank you for your responses. There is much in those responses that I understand already: this is just a weird blind spot I've uncovered in my knowledge, and I'm still not sure I've shaken it. Which is to say, when a signal goes to an ungrounded capacitor, and I take a reading from the negative terminal with my scope, even accepting that I have effectively completed the circuit by connecting my probe to it, why do I get a + and – voltage reading? Why not + and 0?

 
the capacitor will then be charged up to (-) 4.5V.
Why the minus sign in brackets? You seem to be addressing my problem, but I don't quite understand what you mean.
 

Offline Cliff Matthews

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Re: Capacitive coupling
« Reply #15 on: July 15, 2015, 02:01:44 pm »
Any ground difference between the square wave's supply and the scope? Perhaps the scope's on an iso-transformer or its input coupling works better on another channel?
 


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